$K(x,t) = \frac{1}{\sqrt{4\pi t}^N} e^{-\frac{|x|^2}{4t}}$, then $D_x^aD_t^k K(x-y,t)\le C\exp(-c|y|^2)$

Question

Let $$K(x,t) = \frac{1}{\sqrt{4\pi t}^N} e^{-\frac{|x|^2}{4t}}, t>0, x\in\mathbb{R}^n$$

Show that if $u_0\in L^{\infty}(\mathbb{R}^N)$ then $$u(x,t) = \int_{\mathbb{R}^N} u_0(y)K(x-y,t)dy$$ is infinitely differentiable in $\mathbb{R}^N\times ]0,\infty[$.

Suggestion: fix $0<a<b$ and $R>0$ and verify that given $a\in\mathbb{Z}_+^N, k\in\mathbb{Z}_+$, there exists constants $C>0, c>0, p>0$ such that $$D_x^aD_t^k K(x-y,t)\le C\exp(-c|y|^2), x,y\in\mathbb{R}^N, |x|\le R, |y|\le p, t\in [a,b]$$

I've used Leibniz product differentiation rule and ended up with

$$D_t^k K = \sum_i^k{k\choose i}\prod_{j=1}^{k-i}\left(\frac{-N}{2}-j\right)(4\pi t)^{-\frac{N}{2}-j}4^j\pi^i \frac{(-|x|)^{2i}}{4i}e^{-\frac{|x|^2}{4t}}$$

which is already pretty absurd and looks like won't help. I, however, don't see another way of taking thederivative. I know that that product can be converted into a combinatorial but only for even $N$. I still have to take $D_x^a$. Any ideas?

Answer 1

You don't really need to keep track of the coefficients, only that you have some control: $$ D^k_t(K(x,t))=p_k(x,t^{-1}) K(x,t) $$ where $p_k$ is a polynomial of degree $4k$. Similarly, take care of the $D_x^\alpha$ $$ D^\alpha_x(D^k_t(K(x-t)) = Q_{k,\alpha}(x,t^{-1}) K(x,t) $$ with $Q_{k,\alpha}$ a polynomial of degree $4k+2\lvert\alpha\rvert$.

For polynomial $Q_{k,\alpha}$, you can bound \begin{align} \lvert Q_{k,\alpha}(x-y,t)\rvert&\leq \overbrace{(\#\text{monomials})}^{=C(k,\alpha,N)}\times\\ &\quad\times\underbrace{(\max\lvert\text{coefficients}\rvert)}_{\leq C(k,\alpha,N)}\times (\sup\text{bound on monomial}) \end{align} where we use the obvious notation $C(\dots)$ for a universal constant depending only on the those variables listed. Now just bound away each monomial, giving \begin{align} \lvert(x-y)^\beta t^{-j}\rvert&\leq C(R,\lvert\beta\rvert)(1+\lvert y\rvert^2)^{d/2}\cdot\max(1,a^{-j})\\ &\leq C(R,d,a)(1+\lvert y\rvert^2)^{d/2} \end{align} for every degree bound $d$, every $\lvert x\rvert\leq R$, every $y\in\mathbb{R}^N$ and every $t\in[a,b]$. Thus we bound $$ (1+\lvert y\rvert^2)^{d/2}\lvert K(x-y,t)\rvert\leq C(d,R,a,N,\varepsilon)\exp(-(1/(4b)-\varepsilon)\lvert y\rvert^2) $$ for all $\lvert x\rvert\leq R$, $t\in[a,b]$ and $\varepsilon>0$. Note we weaken the coefficient of $\lvert y\rvert^2$ in the exponential to help get the bound in front for all $y\in\mathbb{R}^N$.